Chapter 2 Electric Fields

1 Chapter 2 Electric The Important The Electric FieldSuppose we have a point chargeq0located atrand a set ofexternalcharges conspire so asto exert a forceFon this charge. We can define theelectric fieldat the pointrby:E=Fq0( )The (vector) value of theEfield dependsonlyon the values and locations of the externalcharges, because from Coulomb s law the force on any test charge q0is proportional to thevalue of the charge. However to make this definition really kosher we have to stipulate thatthe test chargeq0is small ; otherwise its presence will significantly influence the locationsof the external Eq. around, we can say that if the Electric field atsome pointrhas the valueEthen asmallcharge placed atrwill experience a forceF=q0E( )The Electric field is avector. From Eq. we can see that its SI units must follows from Coulomb s law that the Electric field at pointrdue to a chargeqlocatedat the origin is given byE=kqr2 r( )where ris the unit vector which points in the same direction Electric Fields from Particular Charge Distributions Electric DipoleAnelectric dipoleis a pair of charges of opposite sign ( q) separated by a distancedwhich is usually meant to be small compared to the distance from the charges at which we1718 Chapter 2.

2 Electric FIELDSEqr(a)Eqr(b)Figure :TheEfield due to a point charge q. (a) If the chargeqis positive, theEfield at some pointa distanceraway has magnitudek|q|/r2and pointsawayfrom the charge. (b) If the chargeqis negative,theEfield has magnitudek|q|/r2and pointstowardthe to find the Electric field. The productqdturns out to be important; the vector whichpoints from the qcharge to the +qcharge and has magnitudeqdis known as theelectricdipole momentfor the pair, and is we form an Electric dipole by placing a charge +qat (0,0, d/2) and a charge qat (0,0, d/2). (So the dipole momentphas magnitudep=qdand points in the +kdirection.) One can show that whenzis much larger thand, the Electric field for points onthezaxis isEz=12 0pz3=k2qdz3( ) Line of ChargeA linear charge distribution is characterized by its charger per unit densityis usually given the symbol ; for an arclengthdsof the distribution, theelectric charge isdq= dsFor a ring of charge with radiusRand total chargeq, for a point on the axis of the ringa distancezfrom the center, the magnitude of the Electric field (which points along thezaxis) isE=qz4 0(z2+r2)3/2( ) Charged Disk & Infinite SheetA two-dimensional (surface) distribution of charge is characterized by its charge per charge densityis usually given the symbol.

3 For an area elementdAof thedistribution, the Electric charge isdq= dAFor a disk or radiusRand uniform charge density on its surface, for a point on the axisof the disk at a distancezaway from the center, the magnitude of the Electric field (whichpoints along thezaxis) isE= 2 0(1 z z2+r2)( ) WORKED EXAMPLES19 The limitR of Eq. gives the magnitude of theEfield at a distancezfrom aninfinite sheet of charge with charge density . The result isE= 2 0( ) Forces on Charges in Electric FieldsAn isolated chargeqin an Electric field experiences a forceF=qE. We note that whenqispositive the force points in the same direction as the field, but whenqis negative, the forceis opposite the field direction!The potential energy of a point charge in anEfield will be discussed at great length inchapter 4!When an Electric dipolepis place in a uniformEfield, it experiences no net force, butitdoesexperience a torque.

4 The torque is given by: =p E( )The potential energy of a dipole also depends on its orientation, and is given by:U= p E( ) Electric Field LinesOftentimes it is useful for us to get anoverall visual pictureof the Electric field due to aparticular distribution of charge. It is useful make a plot where the little arrows represent-ing the direction of the Electric field at each point are joined together, forming continuous(directed) lines . These are theelectric field linesfor the charge a plot will tell us the basicdirectionof the Electric field at all points in space (thoughwe do lose information about themagnitudeof the field when we join the arrows). One canshow that: Electric field lines originate on positive charges (they pointawayfrom the positivecharge) and end on negative charges (they pointtowardthe negative charge). Field lines cannot cross one a diagram of field lines can contain as many lines as you please, for an accuraterepresentation of the field the number of lines originating from a charge should beproportionalto the Worked The Electric Field20 Chapter 2.

5 Electric FIELDSmgFelec = qEEq = 24 mCFigure :Forces acting on the charged mass in Example An object having a net charge of24 Cis placed in a uniform Electric fieldof610 NCdirected vertically. What is the mass of this object if it floats in thefield?[Ser4 23-16]The forces acting on the mass are shown in Fig. The force of gravity points downwardand has magnitudemg(mis the mass of the object) and the electrical force acting on themass has magnitudeF=|q|E, whereqis the charge of the object andEis the magnitudeof the Electric field. The object floats , so the net force is zero. This gives us:|q|E=mgSolve form:m=|q|Eg=(24 10 6C)(610NC)( )= 10 3kgThe mass of the object is 10 3kg = An electron is released from rest in a uniform Electric of the acceleration of the electron. (Ignore gravitation.)[HRW6 23-29]The magnitude of the force on a chargeqin an Electric field is given byF=|qE|, whereEis the magnitude of the field.

6 The magnitude of the electron scharge ise= 10 19C,so the magnitude of the force on the electron isF=|qE|= ( 10 19C)( 104NC) = 10 15 NNewton s 2ndlaw relates the magnitudes of the force and acceleration:F=ma, so theacceleration of the electron has magnitudea=Fm=( 10 15N)( 10 31kg)= 1015ms2 That s themagnitudeof the electron s acceleration. Since the electron has a negative chargethe direction of theforceon the electron (and also the acceleration) isoppositethe directionof the Electric WORKED EXAMPLES21+ +q+qaaP+++Figure :Charge configuration for Example What is the magnitude of a point charge that would create anelectric field maway?[HRW6 23-4]From Eq. , the magnitude of theEfield due to a point chargeqat a distancerisgiven byE=k|q|r2 Here we are givenEandr, so we can solve for|q|:|q|=Er2k=( )( m)2( 109N m2C2)= 10 10 CThemagnitudeof the charge is 10 Calculate the direction and magnitude of the Electric field at pointPinFig.

7 , due to the three point charges.[HRW6 23-12]Since each of the three charges ispositivethey give Electric Fields atPpointingawayfrom the charges. This is shown in Fig. , where the chargesare individually numberedalong with their (vector!)E field note that charges 1 and 2 have the same magnitude and are both at the samedistance fromP. So theE field vectors for these charges shown in Fig. (being in oppositedirections) must cancel. So we are left with only the contribution from charge know the direction for this vector; it is 45 above thexaxis. To find its magnitudewe note that the distance of this charge fromPis half the length of the square s diagonal,or:r=12( 2a) =a 2and so the magnitude isE3=k2qr2=2kq(a/ 2)= 2. Electric Fields + +q+qP+++112233 Figure :Directions for the contributions to theEfield atPdue to the three positive charges inExample 4.

8 +--+ +q+ :Charge configuration for Example the Electric field atPhas magnitudeEnet=4kqa2=4q(4 0)a2=q 0a2and points at an angle of 45 .5. What are the magnitude and direction of the Electric field at the center ofthe square of Fig. ifq= 10 8 Canda= cm?[HRW6 23-13]The center of the square is equidistant from all the distanceris half thediagonal of the square, hencer=12( 2a) =a 2=( cm) 2= 10 2mThen we can find themagnitudesof the contributions to theEfield from each of the charges of magnitudeqhave contributions of ( 109N m2C2)( 10 8C)( 10 2m)2= 104 NCThe charges of magnitude with Fields of twice this magnitude, WORKED EXAMPLES23+--+ +q+ +--+ +q+ +--+ +q+ (a)(b)(c)+--+ +q+ (d)Figure :Directions ofEfield at the center of the square due to three of the corner charges. (a) Upperleft charge is at distancer=a/ 2 from the center (as are the other charges).

9 Efield due to this chargepointsaway fromcharge, in 45 direction. (b)Efield due to upper right charge pointstowardcharge, in+45 direction. (c)Efield due to lower left charge pointstowardcharge, in +225 direction. (d)Efield dueto lower left charge pointsaway fromcharge, in +135 the contributions to the totalEfield are shown in Fig. (a) (d). TheEfield due to the upper left charge pointsaway fromcharge, which is in 45 direction (asmeasured from the +xaxis, as usual. TheEfield due to upper right charge pointstowardthe charge, in +45 direction. TheEfield due to lower left charge pointstowardthat charge,in 180 + 45 = +225 direction. Finally,Efield due to lower right charge pointsaway fromcharge, in 180 45 = +135 we now have the magnitudes and directions of four we add them together?Sure we can!ETotal= ( 104 NC)(cos( 45 )i+ sin( 45 )j)+ ( 105NC)(cos(+45 )i+ sin(45 )j)+ ( 104NC)(cos(225 )i+ sin(225 )j)+ ( 105NC)(cos(+135 )i+ sin(135 )j)(I know, this is the clumsy way of doing it, but I ll get to that.))

10 The sum gives:ETotal= + ( 105NC)jSo the magnitude ofETotalis 105 NCand it points in the + particular problem can be made easier by noting the cancellation of theE s con-tributed by the charges on opposite corners of the square. For example, a +qcharge in theupper left and a + in the lower right is equivalent to asinglecharge +qin thelower right (as far as this problem is concerned). Electric Fields from Particular Charge QuadrupoleFig. shows an Electric quadrupole. It consists of twodipole moments that are equal in magnitude but opposite in direction. Showthat the value ofEon the axis of the quadrupole for points a distancezfrom itscenter (assumez d) is given byE=3Q4 0z4,24 Chapter 2. Electric Fields -q -q+q+qPzdd-p+pFigure :Charges forming Electric quadrupole in Example whichQ(defined byQ 2qd2) is known as thequadrupole momentof the chargedistribution.